2x^2+2y^2+4x+16y+1=0

解答

2 x^{2} + 2 y^{2} + 4 x + 16 y + 1 = 0

(a, b) = (- 1, - 4), r = \frac{33}{2}

求解步骤

2 x^{2} + 2 y^{2} + 4 x + 16 y + 1 = 0

将

2 x^{2} + 2 y^{2} + 4 x + 16 y + 1 = 0

改写为圆标准方程

(x - (- 1))^{2} + (y - (- 4))^{2} = (\frac{33}{2})^{2}

因此圆的性质是:

(a, b) = (- 1, - 4), r = \frac{33}{2}

(x + 89)^{2} + (y + 85)^{2} = (25)^{2}

x^{2} + y^{2} = 4 y + 4

(x - 5)^{2} + (y - 4)^{2} = 1

x^{2} + 2 x + y^{2} + 4 y - 11 = 0

4 x^{2} - 6 x = - 4 y^{2}